Problem
Let $Y_k$ be exponential indipendent random variables of parameter $\alpha k$ for $k\in \{1,2,...,n\}$, i.e. the CDF is $P(Y_k\le x)=1-e^{-\alpha k x}$
Prove that the CDF of $$S_n=\sum_{k=1}^nY_k$$ is $$P(S_n\le y)=\left( 1-e^{-\alpha y}\right)^n$$
My try: Induction
It's obvious for $n=1$. Let's suppose it's true for a certain $n$, then we have (I write "P" for probability, but in some cases I mean probability density):
$$P(S_{n+1}\le y)=P(S_n + Y_{n+1}\le y)= \int_0^{\infty}P(S_n\le y-Y_{n+1}, Y_{n+1}=x)dx = \int_0^{\infty} P(S_n\le y-x| Y_{n+1}=x)P(Y_{n+1}=x)dx$$
Because of the indipendece of the $Y_k$ and knowing the PDF of $Y_{n+1}$ and the CDF of $S_n$ we have
$$P(S_{n+1}\le y)= \int_0^{\infty} \left( 1-e^{-\alpha (y-x)}\right)^n\alpha(n+1)e^{-\alpha(n+1)x} dx = \int_0^{\infty} \alpha(n+1)\sum_{i=0}^n\binom{n}{i}(-1)^ie^{-i\alpha(y-x)}e^{-\alpha(n+1)x}dx = \alpha(n+1)\sum_{i=0}^n \binom{n}{i} e^{-i\alpha y}(-1)^i \int_0^{\infty} e^{-\alpha x(n+1-i)}dx = \alpha(n+1)\sum_{i=0}^n \binom{n}{i} e^{-i\alpha y}(-1)^i \dfrac{1}{\alpha(n+1-i)} = \sum_{i=0}^n (-e^{-\alpha y})^i\dfrac{n+1}{n+1-i}\cdot \dfrac{n!}{i!(n-i)!} = \sum_{i=0}^n (-e^{-\alpha y})^i \binom{n+1}{i}$$
If the sum was from $0$ to $n+1$ this would be correct, but it's not.
What did I do wrong and do you know other ways to prove this theorem?
EDIT: found my mistake, the integral should be from $0$ to $y$ because $$P(S_n\le y)=0$$ if $y<0$.
I would still be curious about a way to solve this problem without induction, i.e. without knowing the solution beforehand.
You don't need to know the solution beforehand to use induction. First compute the distribution of $S_2$ by convolution: $$ f_{S_2}(t) = (f_{Y_1}\star f_{Y_2})(t) = \int_0^t \alpha e^{-\alpha s}2\alpha e^{-2\alpha(t-s)}\ \mathsf ds = 2\alpha e^{-\alpha t}(1-e^{-\alpha t}). $$ Integrating yields $$ F_{S_2}(t) = \int_0^t 2\alpha e^{-\alpha s}(1-e^{-\alpha s)}\ \mathsf ds = (1-e^{-\alpha t})^2. $$ Computing $F_{S_n}(t)$ for a few more values of $n$ would suggest the pattern $F_{S_n}(t) = (1-e^{-\alpha t})^n$, and seeing as this is true for $n=1$, assuming it to be true for some $n\geqslant 1$ we compute the density of $S_{n+1}$ by \begin{align} f_{S_{n+1}}&=(f_{S_n}\star f_{Y_{n+1}})(t)\\ &= \int_0^t n\alpha e^{-\alpha s}(1-e^{-\alpha s})^{n-1}(n+1)\alpha e^{-(n+1)\alpha (t-s)}\ \mathsf ds \\ &=(n+1)\alpha e^{-\alpha t}(1-e^{-\alpha t} )^n. \end{align} Integrating yields $F_{S_{n+1}}(t) = (1-e^{-\alpha t})^{n+1}$, which concludes the proof.